NCERT solutions for class 9 maths chapter 10 Circles Excercise

NCERT solutions for class 9 maths chapter 10 Circles

Excercise: 10.1

Fill in the blanks:

Q1 (i) The centre of a circle lies in _____________ of the circle. (exterior/ interior)

Answer:

The centre of a circle lies in the interior of the circle.

Fill in the blanks:

Q1 (ii) A point, whose distance from the centre of a circle is greater than its radius lies

in_____________ of the circle. (exterior/ interior)

Answer:

A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the

circle.

Fill in the blanks:

Q1 (iii) The longest chord of a circle is a ___________ of the circle.

Answer:

The longest chord of a circle is a diameter of the circle.

Fill in the blanks:

Q1 (iv) An arc is a ________________ when its ends are the ends of a diameter.

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Answer:

An arc is a semi- circle when its ends are the ends of a diameter.

Fill in the blanks:

Q1 (v) Segment of a circle is the region between an arc and ________________ of the circle.

Answer:

Segment of a circle is the region between an arc and chord of the circle.

Fill in the blanks:

Q1 (vi) A circle divides the plane, on which it lies, in _______________ parts.

Answer:

A circle divides the plane, on which it lies, in two parts.

Write True or False: Give reasons for your answers.

Q2 (i) Line segment joining the centre to any point on the circle is a radius of the circle.

Answer:

True. As line segment joining the centre to any point on the circle is a radius of the circle.


Q2 (ii) A circle has only finite number of equal chords.

Answer:

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False . As a circle has infinite number of equal chords.


Q2 (iii) If a circle is divided into three equal arcs, each is a major arc.

Answer:

False. If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas

major arc makes angle greater than 180 degree at centre.


Q2 (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Answer:

True.A chord of a circle, which is twice as long as its radius, is a diameter of the circle.


Q2 (v) Sector is the region between the chord and its corresponding arc.

Answer:

False. As the sector is the region between the radii and arc.


Q2 (vi) A circle is a plane figure.

Answer:

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True. A circle is a plane figure.

Excercise: 10.2

Q1 Recall that two circles are congruent if they have the same radii. Prove that equal chords of

congruent circles subtend equal angles at their centres.

Answer:

Given: The two circles are congruent if they have the same radii.

To prove: The equal chords of congruent circles subtend equal angles at their centres

i.e. BAC= QPR

Proof :

In ABC and PQR,

BC = QR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, ABC PQR (By SSS rule)

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BAC= QPR (CPCT)

Q2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords

are equal.

Answer:

Given : chords of congruent circles subtend equal angles at their centres,

To prove : BC = QR

Proof :

In ABC and PQR,

BAC= QPR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, ABC PQR (By SAS rule)

BC = QR (CPCT)

Excercise: 10.3

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Q1 Draw different pairs of circles. How many points does each pair have in common? What ii

the maximum number of common points?

Answer:

In (i) we do not have any common point.

In (ii) we have 1 common point.

In (iii) we have 1 common point.

In (iv) we have 2 common points.

The maximum number of common points is 2.

Q2 Suppose you are given a circle. Give a construction to find its centre.

Answer:

Given : Points P,Q,R lies on circle.

Construction :

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1. Join PR and QR

2. Draw perpendicular bisector of PR and QR which intersects at point O.

3. Taking O as centre and OP as radius draw a circle.

4. The circle obtained is required.

Q3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector

of the common chord.

Answer:

Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

Construction: Joinpoint P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r) and QM is the bisector of chord AB.

Hence,

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

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Hence,

Now,

PMA and QMA are forming linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

Excercise: 10.4

Q1 Two circles of radii and intersect at two points and the distance between their

centres is . Find the length of the common chord.

Answer:

Given: Two circles of radii and intersect at two points and the distance between their

centres is .

To find the length of the common chord.

Construction: Join OP and draw

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

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Let, PM = x , so QM=4-x

In APM, using Pythagoras theorem

...........................1

Also,

In AQM, using Pythagoras theorem

...........................2

From 1 and 2, we get

Put,x=0 in equation 1

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Q2 If two equal chords of a circle intersect within the circle, prove that the segments of one

chord are equal to corresponding segments of the other chord.

Answer:

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP

= CP and BP=DP.

Construction : Join OP and draw

Proof :

In OMP and ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

OMP = ONP (Both are right angled)

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Thus, OMP ONP (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

......................3

Adding 1 and 3, we have

AM + PM = CN + PN

Subtract 4 from 2, we get

AB-AP = CD - CP

Q3 If two equal chords of a circle intersect within the circle, prove that the line joining the point

of intersection to the centre makes equal angles with the chords.

Answer:

Given: two equal chords of a circle intersect within the circle. Aakas

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To prove: the line joining the point of intersection to the centre makes equal angles with the

chords.

i.e. OPM= OPN

Proof :

Construction: Join OP and draw

In OMP and ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

OMP = ONP (Both are right-angled)

Thus, OMP ONP (By RHS rule)

OPM= OPN (CPCT)

Q4 If a line intersects two concentric circles (circles with the same centre) with centre O at A, B,

C and D, prove that (see Fig. ). Aakas

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Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A,

B, C and D.

To prove : AB = CD

Construction: Draw

Proof :

BC is a chord of the inner circle and

So, BM = CM .................1

(Perpendicular OM bisect BC)

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Similarly,

AD is a chord of the outer circle and

So, AM = DM .................2

(Perpendicular OM bisect AD )

Subtracting 1 from 2, we get

AM-BM = DM - CM

Q5 Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of

radius drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to

Reshma. If the distance between Reshma and Salma and between Salma and Mandip is each,

what is the distance between Reshma and Mandip?

Answer:

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw .

Proof: Aak

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In ORS,

OS = OR and (by construction )

So, RL = LS = 3cm (RS = 6 cm )

In OLS, by pytagoras theorem,

In ORK and OMK,

OR = OM (Radii)

ROK = MOK (Equal chords subtend equal angle at centre)

OK = OK (Common)

ORK OMK (By SAS)

RK = MK (CPCT)

Thus,

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area of ORS = ...............................1

area of ORS = .............................2


Thus,

Q6 A circular park of radius is situated in a colony. Three boys Ankur, Syed and David are

sitting at equal distance on its boundary each having a toy telephone in his hands to talk each

other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

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Construction: AP SD

Proof :

Let AS = SD = AD = 2x cm

In ASD,

AS = AD and AP SD

So, SP = PD = x cm

In OPD, by Pythagoras,

In APD, by Pythagoras,

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Squaring both sides,

Hence, length of string of each phone m

Excercise: 10.5

Q1 In Fig. , A,B and C are three points on a circle with centre O such

that and . If D is a point on the circle other than the arc ABC,

find .

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Answer:

AOC = AOB + BOC=

AOC = 2 ADC (angle subtended by an arc at the centre is double the angle subtended by it

at any)

Q2 A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord

at a point on the minor arc and also at a point on the major arc.

Answer:

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Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and ACB.

Solution :

In OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

ABC is a equilateral triangle.

So, AOB =

AOB = 2 ADB

ACBD is a cyclic quadrilateral .

So, ACB+ ADB =

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Q3 In Fig. , , where P, Q and R are points on a circle with centre O.

Find .

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, PSR + PQR =

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Here, POR = 2 PSR

In OPR ,

OP=OR (Radii )

ORP = OPR (the angles opposite to equal sides)

In OPR ,

OPR+ ORP+ POR=

Q4 In Fig. , find

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Answer:

In ABC,

A+ ABC+ ACB=

A = BDC = (Angles in same segment)

Q5 In Fig. , A, B, C and D are four points on a circle. AC and BD intersect at a point E

such that and . Find

.

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Answer:

DEC+ BEC = (linear pairs)

DEC+ = ( BEC = )

DEC = -

DEC =

In DEC,

D+ DEC+ DCE =

D = BAC (angles in same segment are equal )

BAC =

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Q6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E.

If , is , find . Further, if , find .

Answer:

(angles in the same segment are equal )

In

If AB = BC ,then

Here,

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Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the

quadrilateral, prove that it is a rectangle.

Answer:

AC is the diameter of the circle.

Thus, and ............................1(Angle in a semi-circle is right

angle)

Similarly, BD is the diameter of the circle.

Thus, and ............................2(Angle in a semi-circle is right

angle)


Hence, ABCD is a rectangle.

Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

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Answer:

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of parallelogram )

AD = BC (Given )

so, BE = BC

In EBC,

BE = BC (Proved above )

Thus, ...........1(angles opposite to equal sides )

...............2(Opposite angles of the parallelogram )


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(linear pair)

Thus, ABED is a cyclic quadrilateral.

Q9 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ

are drawn to intersect the circles at A, D and P, Q respectively (see Fig. ). Prove

that .

Answer:

................1(vertically opposite angles)

..................2(Angles in the same segment are equal)

.................3(angles in the same segment are equal)

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From 1,2,3 ,we get

Q10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of

intersection of these circles lie on the third side.

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: AB is the diameter of the circle and ADB is formed in a semi-circle.

ADB = ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and ADC is formed in a semi-circle.

ADC = ........................2(angle in a semi-circle)

From 1 and 2, we have

ADB+ ADC= + =

ADB and ADC are forming a linear pair. So, BDC is a straight line.

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Hence, point D lies on this side.

Q11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that

.

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove :

Proof :

Triangle ABC and ADC are on common base BC and BAC = BDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same

side of line containing line segment, four points lie on the circle.)

CAD = CBD (Angles in same segment are equal)

Q12 Prove that a cyclic parallelogram is a rectangle.

Answer:

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Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

In cyclic quadrilateral ABCD.

.......................1(sum of either pair of opposite angles of a cyclic

quadrilateral)

........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

Excercise: 10.6

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Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two

points of intersection.

Answer:

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : PAQ = PBQ

Proof : In APQ and BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, APQ BPQ (By SSS)

PAQ = PBQ (CPCT)

Q2 Two chords AB and CD of lengths and respectively of a circle are parallel to

each other and are on opposite sides of its centre. If the distance between AB and CD is ,

find the radius of the circle.

Answer:

Given : AB = 5 cm, CD = 11 cm and AB || CD.

To find Radius (OA).

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Construction: Draw

Proof :

Proof: CD is a chord of circle and

Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)

and AN = NB = 2.5 cm

Let OM be x.

So, ON = 6 - x (MN = 6 cm )

In OCM , using Pythagoras,

.............................1

and

In OAN , using Pythagoras,

.............................2

From 1 and 2,

(OC=OA =radii)

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From 2, we get

OA = OC

Thus, the radius of the circle is

Q3 The lengths of two parallel chords of a circle are and . If the smaller chord is at

distance from the centre, what is the distance of the other chord from the centre?

Answer:

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw

Proof :

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Proof: CD is a chord of circle and

Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)

and AN = NB = 4 cm

Let MN be x.

So, ON = 4 - x (MN = 4 cm )

In OCM , using Pythagoras,

.............................1

and

In OAN , using Pythagoras,

.............................2

From 1 and 2,

(OC=OA =radii)

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So, x=1 (since )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle

intersect equal chords AD and CE with the circle. Prove that is equal to half the

difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove :

Construction: Join AC and DE.

Proof :

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Let ADC = x , DOE = y and AOD = z

So, EOC = z (each chord subtends equal angle at centre)

AOC + DOE + AOD + EOC =

.........................................1

In OAD ,

OA = OD (Radii of the circle)

OAD = ODA (angles opposite to equal sides )

OAD + ODA + AOD =

.............................................................2

Similarly,

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.............................................................3

..............................................................4

ODB is exterior of triangle OAD . So,

ODB = OAD + ODA

(from 2)

.................................................................5

similarly,

OBE is exterior of triangle OCE . So,

OBE = OCE + OEC

(from 3)

.................................................................6

From 4,5,6 ;we get

BDE = BED = OEB - OED

..................................................7

In BDE ,

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DBE + BDE + BED =

...................................................8

Here, from equation 1,

...................................9

From 8 and 9,we have

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point

of intersection of its diagonals.

Answer:

Given : ABCD is rhombus.

To prove : the circle drawn with AB as diameter, passes through the point O.

Proof :

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ABCD is rhombus.

Thus, (diagonals of a rhombus bisect each other at )

So, a circle drawn AB as diameter will pass through point O.

Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of

intersection of its diagonals.

Q6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if

necessary) at E. Prove that .

Answer:

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if

necessary) at E.

To prove : AE = AD

Proof :

ADC = 3 , ABC = 4, ADE = 1 and AED = 2

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.................1(linear pair)

....................2(sum of opposite angles of cyclic quadrilateral)

3 = 4 (oppsoite angles of parallelogram )

From 1 and 2,

3+ 1 = 2 + 4

From 3, 1 = 2

From 4, AQB, 1 = 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are

diameters

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof : Aakas

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In ABD and CDO,

AO = OC (Given )

AOB = COD (Vertically opposite angles )

BO = DO (Given )

So, ABD CDO (By SAS)

BAO = DCO (CPCT)

BAO and DCO are alternate angle and are equal .

So, AB || DC ..............1

Also AD || BC ...............2

From 1 and 2,

......................3(sum of opposite angles)

A = C ................................4(Opposite angles of the parallelogram )

From 3 and 4,

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BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a

rectangle.

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

ABCD is a parallelogram. (proved in (i))

(proved in (i))

A parallelogram with one angle , is a rectangle )

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Thus, ABCD is rectangle.

Q8 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F

respectively. Prove that the angles of the triangle DEF

are , and

Answer:

Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F

respectively.

To prove : the angles of the triangle DEF are , and

Proof :

1 and 3 are angles in same segment.therefore,

1 = 3 ................1(angles in same segment are equal )

and 2 = 4 ..................2

Adding 1 and 2,we have

1+ 2= 3+ 4

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,

and

Similarly, and

Q9 Two congruent circles intersect each other at points A and B. Through A any line segment

PAQ is drawn so that P, Q lie on the two circles. Prove that .

Answer:

Given: Two congruent circles intersect each other at points A and B. Through A any line

segment PAQ is drawn so that P, Q lie on the two circles.

To prove : BP = BQ

Proof :

AB is a common chord in both congruent circles.

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In

(Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect,

prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Given :In any triangle ABC, if the angle bisector of and perpendicular bisector of BC

intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

Let ABD = 1 , ADC = 2 , DCB = 3 , CBD = 4

1 and 3 lies in same segment.So,

1 = 3 ..........................1(angles in same segment)

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similarly, 2 = 4 ......................2

also, 1= 2 ..............3(given)

From 1,2,3 , we get

3 = 4

Hence, BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

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NCERT solutions for class 9 maths chapter 10 Circles Excercise

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Transcript of NCERT solutions for class 9 maths chapter 10 Circles Excercise